More math stuff...

[PurplePoot]

Unless I'm doing something wrong here...

sin^2(x) + cos^2(x) = 1

so...

Prove sec^2(x) + csc^2(x) = 1 --

1/sin^2(x) + 1/cos^2(x) = 1

1/(sin^2(x) + cos^2(x)) = 1

1/1 = 1

1 = 1

Thus true.

Now, sub that in

1 = csc^2(x) sec^2(x)

1/sec^2(x) = csc^2(x)

cos^2(x) = csc^2(x)

cos x = csc x

Which is not true.

 

[donut3.5]

Ah, but your mistake was when you add
1/sin^2(x) + 1/cos^2(x), they are not common denominators.
--donut3.5--

 

[PurplePoot]

Ah, haha, got them backwards.

 

[donut3.5]

Ok, proof time.
(csc^2(x))+(sec^2(x))=(csc^2(x))(sec^2(x))
cot^2(x) + 1 + tan^2(x) + 1 = (csc^2(x))(sec^2(x)) (Pythagorean ID)
cos^2(x)/sin^2(x) + sin^2(x)/cos^2(x) + 2 = (csc^2(x))(sec^2(x)) (quotient ID)
((cos^2(x)+sin^2(x))^2)/sin^2(x)cos^2(x)= (csc^2(x))(sec^2(x)) (lots of multiplication, adding, and whatever)
After lots of factoring that I decided to not take notes on, you get
1/(cos^2(x)sin^2(x)) = (csc^2(x))(sec^2(x)), which is true by reciprocal ID.

If anyone is still interested I can probably try and remember how I got those middle steps.
--donut3.5--

 

[Eleandor]

I think you're right. I also don't see how you came to this topic if not expecting such a thing

I like sin/cos

Donut3.5, unless I'm doing something I can't do, isn't it MUCH simpler if you did this:

(csc^2(x))+(sec^2(x))=(csc^2(x))(sec^2(x)) Removing the x to make it easier to read
==> (1/sin²) + (1/cos²) = (1/sin²) * (1/cos²) converting sec / cosec into 1/cos and 1/sin
==> (sin² + cos²)/(sin²*cos²) = 1/(sin²*cos²) sin²+cos² = 1, then removing the sin²*cos² in left/right
==> 1 = 1?

 

[donut3.5]

When solving for proofs, you can't do stuff to both sides, you have to pick and edit only one.
I just realized what I wrote wrong on my previous post, I'll edit it.
--donut3.5--

 

[Eleandor]

Yes you can...
If you say A is equal to B, one way to prove it is that you do stuff with both A and B (things that you can do, e.g. you can't say that -2 = 2 since 4 = 4) until you can clearly see A = B. My proof is therefor a right one.

 

[donut3.5]

Actually I found an even easier one. (And btw, math teachers prefer if you stick to one side, iunno why )
(csc^2(x))+(sec^2(x))=(csc^2(x))(sec^2(x)) given
(1/sin^2(x))+1/cos^2(x))=(csc^2(x))(sec^2(x)) reciprocal ID
((sin^2(x)+cos^2(x))/(sin^2(x)cos^2(x))=(csc^2(x))(sec^2(x)) Get common denominators and add
then you get
1/(sin^2(x)cos^2(x))=(csc^2(x))(sec^2(x)) by Pythagorean ID
Then reciprocal gets you...
(csc^2(x))(sec^2(x))=(csc^2(x))(sec^2(x))

I think that's a really similar way to what you did Eleandor (I mean elenai ) but modifying only one side.
--donut3.5--

 

[GST_Nemisis]

I think people who find sin/cos stuff awesome are nerds...

pro quote.
. .

 

[donut3.5]

I think you're right. I also don't see how you came to this topic if not expecting such a thing

pro-er quote
--donut3.5--